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#### The ratio of the corresponding altitudes of two similar triangles is 3/5. Is it correct to say that ratio of their areas is 6/5? Why?

Given:- Ratio of corresponding altitudes of two similar triangles is 3/5

As we know that the ratio of the areas of two similar triangles is equal to the ratio of squares of any two corresponding altitudes.

$\therefore \frac{\text {Area 1}}{\text {Area 2}}=\left ( \frac{\text {Altitude 1}}{\text {Altitude 2}} \right )^{2}$

$\frac{\text {Area 1}}{\text {Area 2}}=\left ( \frac{3}{5} \right )^{2}$            $\left [ Q\frac{\text {Altitude 1}}{\text {Altitude 2}}=\frac{3}{5} \right ]$

$=\frac{9}{25}$

Hence the given statement is false because ratio of areas of two triangles is 9/25 which is not equal to 6/5

#### If in two right triangles, one of the acute angles of one triangle is equal to an acute angle of the other triangle, can you say that the two triangles will be similar? Why?

Let two right angle triangles are ABC and PQR.

Given:- One of the acute angle of one triangle is equal to an acute angle of the other triangle.

$\text {In}\Delta ABC \; \text {and}\; \Delta PQR$

$\\\angle B = \angle Q = 90^{o}\\\angle C = \angle R = x^{o}$

In $\Delta ABC$

$\angle A + \angle B + \angle C = 180^{o}$

[Sum of interior angles of a triangle is 180°]

$\angle A + 90^{o}+ x^{o} = 180^{o}$

$\angle A + 90^{o}- x^{o} \; \; \; \; \; \; \; \; \; ....(1)$

Also in $\Delta PQR$

$\angle P + \angle Q + \angle R = 180^{o}$

$\angle P + 90^{o}- x^{o} = 180^{o}$

$\angle P = 90^{o}- x^{o} \; \; \; \; \; \; .....(2)$

from equation (1) and (2)

$\angle A=\angle P \; \; \; \; \; \; .....(3)$

from equation (1), (2) and (3) are observed that corresponding angles of the triangles are equal therefore triangle are similar.

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#### D is a point on side QR of $\Delta$PQR such that PD $\perp$ QR. Will it be correct to say that $\Delta$PQD ~ $\Delta$RPD? Why?

In $\Delta$PQD and $\Delta$RPD

$PD = PD$          (common side)

$\angle PDQ = \angle PDR$            (each 90o)

The given statement $\Delta PQD \sim \Delta RPD$ is false.

Because according to the definition of similarity two triangles are similar, if their corresponding angles are equal and their corresponding sides are in the same ratio

#### In Fig., if $\angle$D = $\angle$C, then is it true that $\Delta ADE \sim \Delta ACB$ ? Why?

In $\Delta$ ADE and $\Delta$ ACB

$\angle D=\angle C$        (given)

$\angle A = \angle A$      (common angle)

And we know that if two angles of one triangle are equal to the two angles of another triangle, then the two triangles are similar by AA similarity criteria.

$\therefore$   $\Delta ADE \sim \Delta ACB$         {by AA similarity criterion}

Hence the given statement is true.

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#### Is it true to say that if in two triangles, an angle of one triangle is equal to an angle of another triangle and two sides of one triangle are proportional to the two sides of the other triangle, then the triangles are similar? Give reasons for your answer.

Given:- (1) An angle of one triangle is equal to an angle of another triangle.

(2) Two sides of one triangle are proportional to the two sides of the other triangle.

According to SAS similarity criterion if one angle of a triangle is equal to one angle of another triangle and the sides including these angles are in the same ratio, then the triangles are similar.

But here, the given statement is false because one angle and two sides of two triangles are equal but these sides not including equal angle.

#### Two sides and the perimeter of one triangle are respectively three times the corresponding sides and the perimeter of the other triangle. Are the two triangles similar? Why?

According to question,

$AB = 3PQ \; \; \; \; \; \; ...(1)$

$AC = 3PR \; \; \; \; \;...(2)$

also the perimeter of $\Delta$ABC is three times the perimeter of $\Delta$PQR

$AB + BC + CA = 3(PQ + QR + RP)$

$AB + BC + CA = 3PQ + 3QR + 3RP$

$3PQ + BC + 3PR = 3PQ + 3QR + 3PR$                   (using eq. (1) and (2))

$BC=3QR\; \; \; \; \; \; .....(3)$

from equation (1), (2) and (3) we conclude that sides of both the triangles are in the same ratio.

As we know that if the corresponding sides of two triangles are in the same ratio, then the triangles are similar by SSS similarity criterion

Hence given statement is true.

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#### Is the following statement true? Why? “Two quadrilaterals are similar, if their corresponding angles are equal”.

Quadrilateral:- A quadrilateral can be defined as a closed, two-dimensional shape which has four straight sides. We can find the shape of quadrilaterals in various things around us, like in a chessboard, a kite etc. The given statement "Two quadrilaterals are similar if their corresponding angles are equal" is not true because. Two quadrilaterals are similar if and only if their corresponding angles are equal and the ratio of their corresponding sides are also equal.

#### In triangles PQR and MST, $\angle$P = 55°, $\angle$Q = 25°, $\angle$M = 100° and $\angle$S = 25°. Is $\Delta$QPR $\sim$ $\Delta$TSM? Why?

$\Delta PQR, \angle P + \angle Q + \angle R = 180^{o}$$\text{ ({Interior angle of triangle}) }$

$55+25+\angle R=180$

$\angle R=180-80$

$\angle R=100^{o}$

$\text{In }$  $\Delta TSM, \angle T + \angle S + \angle M = 180^{o}$$\text{[Interior angle of the triangle] }$

$\angle T + 25^{o}+100^{o}=180^{o}$

$\angle T =180^{o}-125$

$\angle T =55^{o}$

$\text{In }$  $\Delta PQR$ $\text{and }$ $\Delta TSM$

$\angle P=\angle T$

$\angle Q=\angle S$

$\angle R=\angle M$

Also are know that if all corresponding angles of two triangles are equal then triangles are similar.

$\therefore \Delta PQR\sim \Delta TSM$

Hence given statement is false because $\angle$QPR is not similar to $\angle$TSM.

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#### In Fig 6.4, BD and CE intersect each other at the point P. Is $\Delta$PBC $\sim$ $\Delta$PDE? Why?

Given:- BD and CE intersect each other at point P.

$\text{Here, }$  $\angle BPC=\angle EPD$ $\text{[Vertically opposite angle] }$

$\frac{BP}{PD}=\frac{5}{10}=\frac{1}{2}$

$\frac{PC}{PE}=\frac{6}{12}=\frac{1}{2}$

$\frac{BP}{PD}=\frac{PC}{PE}$

Two triangles are similar if their corresponding angles are equal and their corresponding sides are in the same ratio.

$\text{Also in }$ $\Delta PBC$ $\text{ and }$$\Delta PDE$

$\frac{BP}{PD}=\frac{PC}{PE}\; and\; \angle BPC=\angle EPD$

$\therefore \Delta PBC\sim \Delta PDE$

Hence given statement is true.

#### A and B are respectively the points on the sides PQ and PR of a triangle PQR such that PQ = 12.5 cm, PA = 5 cm, BR= 6 cm and PB = 4 cm. Is AB || QR?Give reasons for your answer.

Given: PQ = 12.5 cm, PA = 5 cm, BR = 6 cm and PB = 4 cm

QA = QP – PA = 12.5 – 5 = 7.5 cm

$\frac{PA}{AQ}=\frac{5}{7.5}=\frac{50}{75}=\frac{2}{3}$

$\frac{PB}{PR}=\frac{4}{6}=\frac{2}{3}$

Here

$\frac{PA}{AQ}=\frac{PB}{PR}$

According to Converse of basic proportionality theorem if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

$\therefore AB\parallel QR$